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October 24

Matroid morphism

How do you define matroid homomorphism (matroid map, morphism of matroids)? 212.87.13.70 (talk) 10:09, 24 October 2008 (UTC)[reply]

There are two candidates: strong map relation and weak map relation. The latter is defined by:

The inverse image of any independent set is an independent set.

For strong map relation see http://www.sciencedirect.com/science/article/pii/S0012365X05004231

Help with problem

I need HELP: 3^(2x+1)+(28*3^(x))+9=0 —Preceding unsigned comment added by 72.205.199.12 (talk) 23:09, 24 October 2008 (UTC)[reply]

Rewrite

3^{2x+1}+28*3^{x}+9=0

as

3^{2x+1}+3^{3x}+3^{x}+3^{2}=0

Mdob | Talk 00:04, 25 October 2008 (UTC)[reply]

Could you explain? —Preceding unsigned comment added by 72.205.199.12 (talk) 00:21, 25 October 2008 (UTC)[reply]

Mdob, do you mean

3^{2x+1}+3^{3+x}+3^{x}+3^{2}=0

(note, 3+x not 3x)? --Tango (talk) 00:29, 25 October 2008 (UTC)[reply]

Ooops! Yeah, I botched the equation....:P Thanks again, Tango.

There seems to be something odd with this equation... if we extract the logarithm in base 3 of the lhs we get a logarithm of zero in the rhs. What am I doing wrong? Mdob | Talk 01:02, 25 October 2008 (UTC)[reply]

It makes no sense at all to take logarithms, since on the left side you get a logarithm of a sum. Please. Don't be that clumsy. Michael Hardy (talk) 02:31, 25 October 2008 (UTC)[reply]

Doh! You are correct, of course. mea culpa. thanks. Mdob | Talk 11:34, 25 October 2008 (UTC)[reply]

Mdob, you seem to think 3³ = 28. That is incorrect. Michael Hardy (talk) 02:05, 25 October 2008 (UTC)[reply]

Uh?

27*3^{x}+3^{x}=28*3^{x}

, I've checked this with the Eigenmath CAS, so my expansion of the middle term is correct. Mdob | Talk 11:34, 25 October 2008 (UTC)[reply]

but the rest of your solution is correct, of course. Mdob | Talk 11:34, 25 October 2008 (UTC)[reply]

There are no solutions that are real numbers, since if w is real then 3^w is positive, so you'd be adding positive numbers and getting zero, which is impossible. However, supposing u = 3^x, then we have:

{\begin{aligned}3^{2x+1}+28\cdot 3^{x}+9&=0\\(3^{x})^{2}\cdot 3+28\cdot 3^{x}+9&=0\\3u^{2}+28u+9&=0,\end{aligned}}

and this is a quadratic equation. Plugging the three coefficients into the quadratic formula and simplifying, we get

u={\frac {-1}{3}}{\text{ or }}u=-9.

So we need

3^{x}=u=-9{\text{ or }}3^{x}=u={\frac {-1}{3}}.

As I said, if x is a real number, then 3^x is positive, so let us seek solutions that are non-real complex numbers. Now

3^{x}=e^{x\ln 3}.\,

Now suppose u and v are the real and imaginary parts of x respectively. Then (OK, at this point I'm really wondering if maybe you misread 3x² − 28x + 9 = 0, since that would work out much more neatly, with real numbers):

3^{x}=e^{(u+iv)\ln 3}=e^{u\ln 3}(\cos(v\ln 3)+i\sin(v\ln 3)=3^{u}(\cos(v\ln 3)+i\sin(v\ln 3).\,

So you'll need 3^u = 9, cos = −1, and sin = 0. That gives us u = 2, and v ln 3 = π ± 2πn., etc. Michael Hardy (talk) 02:26, 25 October 2008 (UTC)[reply]

Thanks! —Preceding unsigned comment added by 72.205.199.12 (talk) 17:45, 25 October 2008 (UTC)[reply]